Respuesta :

From the secant theorem we know that:

[tex] 3\cdot(3+k)=5\cdot(5+6)\\\\9+3k=5\cdot11\\\\9+3k=55\\\\3k=55-9\\\\3k=46\qquad|:3\\\\\boxed{k=15\frac{1}{3}} [/tex]

6y6

3(k+3)=5(5+6)

3k+9=55

3k=46

k = [tex] \frac{46}{3} [/tex] = 15 1/3